In an AP, (a_4+a_9=95) and (a_{16}=137). What is (a_{31})?
Answer and explanation
Correct answer: \(\frac{5288}{19}\)
Let the first term be \(a\) and the common difference be \(d\). Then \(a_4+a_9=(a+3d)+(a+8d)=2a+11d=95\), while \(a_{16}=a+15d=137\). Doubling the second equation and subtracting the first gives \(19d=179\), so \(d=\frac{179}{19}\). Therefore, \(a_{31}=a_{16}+15d=137+15\left(\frac{179}{19}\right)=\frac{5288}{19}\). The value \(263\) would require \(d=9\), which does not satisfy both given conditions. Exam tip: Use \(a_{31}=a_{16}+15d\) to calculate a distant term efficiently.
Frequently asked questions
What is the correct answer to this question?
\(\frac{5288}{19}\)
Why is this the correct answer?
Let the first term be \(a\) and the common difference be \(d\). Then \(a_4+a_9=(a+3d)+(a+8d)=2a+11d=95\), while \(a_{16}=a+15d=137\). Doubling the second equation and subtracting the first gives \(19d=179\), so \(d=\frac{179}{19}\). Therefore, \(a_{31}=a_{16}+15d=137+15\left(\frac{179}{19}\right)=\frac{5288}{19}\). The value \(263\) would require \(d=9\), which does not satisfy both given conditions. Exam tip: Use \(a_{31}=a_{16}+15d\) to calculate a distant term efficiently.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Arithmetic Progressions (AP). Topic: Finding the $n$th term of an AP.
